In the case of a normal distribution, there is no finite range that contains 100% of the values, hence, the range containing ______ of the values is taken as the benchmark.
Skill/Topic: Statistical Process ControlA) 99.73%B) 99.99%C) 99.10%D) 99.78%
Skill/Topic: Statistical Process ControlA) 4B) 5C) 6D) 8Explanation: Process Capability = 6σ
Latest Answer: Process Capability is 6 times standard deviationsudhakar kolla,Email ID:kollasudhakar2005@yahoo.co.in.Ph:+91-9819859346 ...
The process capability ratio (PCR, or Cp) compares the tolerance specified on the characteristic with the process capability
Skill/Topic: Statistical Process ControlA) TrueB) False
Skill/Topic: Statistical Process ControlA) TrueB) False
Skill/Topic: Statistical Process ControlA) Cp = (USL-LCL)/ 6σExplanation: In this formula, USL and LSL are the upper and lower specification limits of the quality characteristic, respectively. The
Latest Answer: Cp = (USL-LCL)/ 6σExplanation: In this formula, USL and LSL are the upper and lower specification limits of the quality characteristic, respectively. The quantity in the denominator, 6σ signifiesthe fact that 99.73% of the values generated by ...
The specification limits for the inside diameter of a hole are (0.995", 1.005"). The standard deviation of the inside diameters generated by a lathe machine selected to process this component is estimated to be 0.002". Compute the Cp index of this process. Assume that the inside diameters generated by the machine follow normal distribution
Skill/Topic: Statistical Process ControlA) 0.7444B) 0.8333C) 0.9673D) 0.8777Explanation: Cp = (USL-LCL)/ 6σ = 1.005 - 0.995 / (6 * 0.002 )= 0.8333
Skill/Topic: Statistical Process ControlA) TrueB) FalseExplanation: In this formula, P[Z> 3Cp ] can be obtained from the Standard Normal Distribution Tables
Latest Answer: True ...
For the example given in Question 26, estimate the proportion of defectives if P[Z> 3Cp ] i.e. P[Z> 3* 0.8333 ] = P[ Z> 2.4999] from the Normal table is found as 0.0062
Skill/Topic: Statistical Process ControlA) 0.0122
The roughness of the ground surface of a component cannot exceed 0.02 units. A random sample of components ground by a surface-grinding machine yielded the following estimates: Mean roughness = 0.01, Standard deviation = 0.003. Compute the Cp index of this process
Skill/Topic: Statistical Process ControlA) 1.111Explanation: Cp = [USL – μ]/ 3σ = [0.02 - 0.01]/[3 * 0.003] = 1.111
The tensile strength of a welded joint used in construction has to be at least equal to 40 tons per square inch (ton/in.2). A random sample of 100 joints welded by a welding machine yielded the following estimates of the parameters of the distribution of the tensile strengths: Mean = 65 ton/in2. Standard deviation = 8.2 ton/ in2..
Skill/Topic: Statistical Process ControlA) 1.012Explanation: Cp = [μ- LSL]/ 3σ = [65- 40]/[3*8.2] = 1.012
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